SOLUTION: Find three consecutive odd integers such that the square of the second added to the first is seventy five less than the square of the third.

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Question 314693: Find three consecutive odd integers such that the square of the second added to the first is seventy five less than the square of the third.
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
Let x, x+2 & x+4 be the 3 integers.
(x+2)^2+x=(x+4)^2-75
x^2+4x+4+x=x^2+8x+16-75
x^2-x^2+5x-8x=-4+16-75
-3x=-63
x=-63/-3
x=21 ans.
Proof:
(21+2)^2+21=(21+4)^2-75
23^2+21=25^2-75
529+21=625-75
550=550